Four equal resistance dissipated 5 W of power together when connected in series to a battery of negligible internal resistance . The total power dissipated in these resistance when connected in parallel across the same battery would be .

Options

A. 125W
B. 80W
C. 20W
D. 5W

Answer

To solve the problem step by step, we will first analyze the situation when the resistances are connected in series and then when they are connected in parallel.

### Step 1: Determine the equivalent resistance in series
When four equal resistances \( R \) are connected in series, the total or equivalent resistance \( R_1 \) can be calculated as:
\[ R_1 = R + R + R + R = 4R \]

**Hint:** Remember that in a series circuit, the total resistance is the sum of all individual resistances.

### Step 2: Use the power formula for the series connection
The power \( P_1 \) dissipated in the series connection is given by the formula:
\[ P_1 = \frac{V^2}{R_1} \]
Substituting \( R_1 \) into the equation:
\[ P_1 = \frac{V^2}{4R} \]
We know from the problem statement that \( P_1 = 5 \, \text{W} \):
\[ \frac{V^2}{4R} = 5 \]
From this, we can rearrange to find \( V^2 \):
\[ V^2 = 20R \quad \text{(Equation 1)} \]

**Hint:** This equation relates the voltage \( V \) to the resistance \( R \) based on the power dissipated.

### Step 3: Determine the equivalent resistance in parallel
When the same four resistances are connected in parallel, the equivalent resistance \( R_2 \) can be calculated using the formula for parallel resistances:
\[ \frac{1}{R_2} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{4}{R} \]
Thus, simplifying gives:
\[ R_2 = \frac{R}{4} \]

**Hint:** In parallel circuits, the reciprocal of the total resistance is the sum of the reciprocals of the individual resistances.

### Step 4: Use the power formula for the parallel connection
The power \( P_2 \) dissipated in the parallel connection can be expressed as:
\[ P_2 = \frac{V^2}{R_2} \]
Substituting \( R_2 \) into the equation:
\[ P_2 = \frac{V^2}{R/4} = \frac{4V^2}{R} \]
Now, substitute \( V^2 \) from Equation 1:
\[ P_2 = \frac{4 \times 20R}{R} = 80 \, \text{W} \]

**Hint:** Make sure to substitute the correct value for \( V^2 \) to find the power in the parallel configuration.

### Final Answer
The total power dissipated in these resistances when connected in parallel across the same battery is:
\[ \boxed{80 \, \text{W}} \]